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Sep 3 2016 07:32pm
Integral [(1 - 2x) / (x^2 + x +3)] dx
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Sep 3 2016 08:12pm
i've been out too long. i dont remember any of the trig integrals personally.

Code
http://www.wolframalpha.com/input/?i=integrate+%5B(1+-+2x)+%2F+(x%5E2+%2B+x+%2B3)%5D+dx


This post was edited by carteblanche on Sep 3 2016 08:12pm
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Sep 3 2016 08:40pm
Quote (carteblanche @ Sep 3 2016 09:12pm)
i've been out too long. i dont remember any of the trig integrals personally.

Code
http://www.wolframalpha.com/input/?i=integrate+%5B(1+-+2x)+%2F+(x%5E2+%2B+x+%2B3)%5D+dx


how can u tell its trig integrals without lookin at solutions?

dude, that looks nasty to solve lol wtf
btw this is just the right side of the differential equation that I have to solve... the left side of the Differntial equation is just Integral dx/x

This post was edited by FamilyGuyViewer on Sep 3 2016 08:44pm
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Sep 3 2016 09:09pm
Looks like you'll need arctan and ln
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Sep 3 2016 09:44pm
Quote (FamilyGuyViewer @ Sep 3 2016 10:40pm)
how can u tell its trig integrals without lookin at solutions?


didn't you learn a chart in calc 1?



im too old for this. if it was a simple u substitution i would have done it (eg if top was 1+2x instead of 1-2x).

This post was edited by carteblanche on Sep 3 2016 09:45pm
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Sep 4 2016 07:12am
Quote (carteblanche @ Sep 3 2016 10:44pm)
didn't you learn a chart in calc 1?

http://www.drcruzan.com/Images/Mathematics/IndefiniteIntegral/InverseTrigIntegralsTable.png

im too old for this. if it was a simple u substitution i would have done it (eg if top was 1+2x instead of 1-2x).


that was years ago lol

so how the fck am i suppose to do this by hand

This post was edited by FamilyGuyViewer on Sep 4 2016 07:12am
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Sep 4 2016 02:21pm
Quote (FamilyGuyViewer @ Sep 4 2016 09:12am)
that was years ago lol

so how the fck am i suppose to do this by hand


have you tried integration by parts?
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Sep 4 2016 02:43pm
(1-2x) / (x² + x +3) = - (2x + 1) / (x² + x + 3) + 2 / (x² + x + 3)

Integrate the first term simply by using ∫ u'/u = ln |u|

(don't forget to show that the denominator is always positive).

For the second term, show that : x² + x + 3 = 11/4 [ ((2x + 1)/√11)² + 1]

(completing the square)

and change the variable :
t = (2x + 1) / √11
dt = 2 dx / √11

and you should recognize ∫ dt / ( t² + 1) , which is Arctan t.
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Sep 4 2016 03:43pm
Quote (feanur @ Sep 4 2016 03:43pm)
(1-2x) / (x² + x +3) = - (2x + 1) / (x² + x + 3) + 2 / (x² + x + 3)

Integrate the first term simply by using ∫ u'/u = ln |u|

(don't forget to show that the denominator is always positive).

For the second term, show that : x² + x + 3 = 11/4 [ ((2x + 1)/√11)² + 1]

(completing the square)

and change the variable :
t = (2x + 1) / √11
dt = 2 dx / √11

and you should recognize ∫ dt / ( t² + 1) , which is Arctan t.


thanks alot man. dam i woulda been stuck on this problem for forever if this was on my exam...
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Sep 4 2016 03:46pm
If this is not for an exam, just ask Wolfram ;)
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